In a distribution exactly normal 10.03% of the
items are under 25 kg wt and 89.97% of the items
are under 70 kg wt. What are the mean and S.D of
the distribution?
Answers
Given : distribution exactly normal 10.03% of the items are under 25 kg wt and 89.97% of the items are under 70 kg wt.
To Find : mean and S.D of the distribution?
Solution:
Let say Mean = M
& SD = S
Z score = ( Value - Mean)/SD
10.03% of the items are under 25 kg wt
10.03% corresponds to Z score = - 1.28
=> -1.28 = ( 25 - M)/S
=> -12.8S = 25 - M
89.97% of the items are under 75 kg wt
89.97% corresponds to Z score = 1.28
=> 1.28 = ( 70 - M)/S
=> 12.8S = 70 - M
Adding Both 0 =95 - 2M
=> M = 47.5
Mean Weight = 47.5 kg
12.8S = 70 - M
=> 1.28S = 70 - 47.5
=> 1.28S = 22.5
=> S = 17.58
mean and S.D of the distribution are 47.5 and 17.58 respectively
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Answer:
M = 47.5
S = 17.58
Mean and S.D of the distribution are 47.5 and 17.58, respectively
Step-by-step explanation:
Given: distribution exactly normal 10.03% of the items are under 25 kg wt, and 89.97% of the items are under 70 kg wt.
To Find: mean and S.D of the distribution?
Solution:
Let's say Mean = M
& SD = S
Z score = ( Value - Mean)/SD
10.03% of the items are under 25 kg wt
10.03% corresponds to Z score = - 1.28
=> -1.28 = ( 25 - M)/S
=> -12.8S = 25 - M
89.97% of the items are under 75 kg wt
89.97% corresponds to Z score = 1.28
=> 1.28 = ( 70 - M)/S
=> 12.8S = 70 - M
Adding Both 0 =95 - 2M
=> M = 47.5
Mean Weight = 47.5 kg
12.8S = 70 - M
=> 1.28S = 70 - 47.5
=> 1.28S = 22.5
=> S = 17.58
mean and S.D of the distribution are 47.5 and 17.58, respectively